The uniform distribution#

In Example: Uncertainties in steel ball manufacturing, we had a random variable for which all values between a given interval were equally probable. This is a very common situation covered by the so-called uniform distribution.

Let’s start with the simplest case: a random variable taking values between 0 and 1 with constant probability density. We write:

and we read

$X$ follows a uniform distribution taking values in $[0,1]$.

The probability density of the uniform is constant in $[0,1]$ and zero outside it. We have:

What should the constant $c$ be? Just like in Example: Uncertainties in steel ball manufacturing, you can find it by ensuring the PDF integrates to one (see PDF Property 5:

So, the PDF is:

To find the CDF, we can use PDF Property 3: $$F(x)=p(X\le x)={\int }_0^{x}p(\tilde{x})d\tilde{x}={\int }_0^{x}d\tilde{x}=x.$$Obviously,wehave$F(x) = 0$for$x < 0$and$F(x) = 1$for$x > 1$.

Using this result, we can find the probability that $X$ takes values in $[a,b]$ for $a<b$ in $[0,1]$. It is: $$p(a\le X\le b)=F(b)-F(a)=b-a.$$

Instantiating the uniform using scipy.stats#

Let me know show you how you can make a uniform random variable using scipy:

Show code cell source Hide code cell source

import matplotlib.pyplot as plt
%matplotlib inline
import seaborn as sns
sns.set(rc={"figure.dpi":100, 'savefig.dpi':300})
sns.set_context('notebook')
sns.set_style("ticks")
from IPython.display import set_matplotlib_formats
set_matplotlib_formats('retina', 'svg')
import numpy as np
import scipy.stats as st

X = st.uniform()

Here is how you can get some samples:

X.rvs(size=10)

array([0.33842053, 0.57010738, 0.1323871 , 0.58357996, 0.68875828,
       0.67923023, 0.84656861, 0.75593012, 0.25791332, 0.13214875])

You can evalute the PDF at any point like this:

X.pdf(0.5)

1.0

X.pdf(-0.1)

0.0

X.pdf(1.5)

0.0

Let’s plot the PDF:

fig, ax = plt.subplots()
xs = np.linspace(-0.1, 1.1, 200)
ax.plot(xs, X.pdf(xs))
ax.set_xlabel('$x$')
ax.set_ylabel('$p(x)$');

You can evaluate the CDF like this:

X.cdf(-0.5)

0.0

X.cdf(0.5)

0.5

X.cdf(1.2)

1.0

Let’s plot the CDF:

fig, ax = plt.subplots()
ax.plot(xs, X.cdf(xs))
ax.set_xlabel('$x$')
ax.set_ylabel('$F(x)$');

Finally, let’s find the probability that $X$ is between two numbers $a$ and $b$. For the uniform it is, of course, trivial, but let’s see how it is done using the scipy functionality:

a = -1.0
b = 0.3
prob_X_is_in_ab = X.cdf(b) - X.cdf(a)
print('p({0:1.2f} <= X <= {1:1.2f}) = {2:1.2f}'.format(a, b, prob_X_is_in_ab))

p(-1.00 <= X <= 0.30) = 0.30

The uniform distribution over an arbitrary interval $[a,b]$#

The uniform distribution can also be defined over an arbitrary interval $[a,b]$. We write: $$X\sim U([a,b]).$$ We read:

$X$ follows a uniform distribution on $[a,b]$.

The PDF of this random variable is:

where $c$ is a positive constant. This simply tells us that the probability density of finding $X$ in $[a,b]$ is something positive and that the probability density of findinig outside is exactly zero. This is exactly the situation we had in Example: Uncertainties in steel ball manufacturing. The positive constant $c$ is determined by imposing the normalization condition:

This gives:

From this we get:

and we can now write:

From the PDF, we can now find the CDF for $x\in [a,b]$:

Instantiating the generic uniform using scipy.stats:#

Let’s instantiate using scipy.stats:

a = -2.0
b = 5.0
X = st.uniform(loc=a, scale=(b-a))

Some samples:

X.rvs(size=10)

array([-1.63839392, -1.08615066, -1.9249188 , -0.86574048, -0.01377033,
        0.91441   ,  2.94486672, -0.90602192, -0.44926847,  4.28172985])

The PDF:

fig, ax = plt.subplots()
xs = np.linspace(a - 0.2, b + 0.2, 200)
ax.plot(xs, X.pdf(xs))
ax.set_xlabel('$x$')
ax.set_ylabel('$p(x)$');

The CDF:

fig, ax = plt.subplots()
ax.plot(xs, X.cdf(xs))
ax.set_xlabel('$x$')
ax.set_ylabel('$F(x)$');

Questions#

• Repeat the code above so that the random variable is $U([1,10])$.